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Math Help - Sum to infinity

  1. #1
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    Sum to infinity


    Could someone please explain how to solve this problem cheers.
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  2. #2
    MHF Contributor Mathstud28's Avatar
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    Quote Originally Posted by Haris View Post

    Could someone please explain how to solve this problem cheers.
    1-x^2+x^4-x^6+\cdots=\sum_{n=0}^{\infty}(-1)^nx^{2n}

    The radius of convergence can be found using the Root test...we must have that \lim_{n\to\infty}\sqrt[n]{|(-1)^nx^{2n}|}=\lim_{n\to\infty}x^2<1

    And where the series converges we can use the resmblance to the geometric series to obtain that \sum_{n=0}^{\infty}(-1)^nx^{2n}=\sum_{n=0}^{\infty}\left(-x^2\right)^n=\frac{1}{1-(-x^2)}=\frac{1}{1+x^2}
    Last edited by Mathstud28; December 14th 2008 at 11:25 AM.
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  3. #3
    Math Engineering Student
    Krizalid's Avatar
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    1-x^{2}+x^{4}-x^{6}+\cdots +(-1)^nx^{2n}+\cdots=\sum\limits_{n=0}^{\infty }{(-1)^{n}x^{2n}}.

    Now put a_{n}=(-1)^{n}x^{2n} then \underset{n\to \infty }{\mathop{\lim }}\,\sqrt[n]{\left| a_{n} \right|}=\underset{n\to \infty }{\mathop{\lim }}\,x^{2}=x^{2} and the convergence requires that x^2<1\implies-1<x<1.
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